Calculus
posted by Jeff .
If a_n does not equal zero for any n>=1 and ∑a_n converges absolutely, then ∑ 1/a_n diverges. The series are from n=1 to infinity.
I think this is true but I'm not sure.

If the series converges, then the terms must approach zero. In fact, all terms after the Nth (for some N) must be less than a, for some small a < 1.
So, since the terms are approaching zero, their reciprocals get larger and larger  and there are infinitely many of them ...
Respond to this Question
Similar Questions

math
Write the arithmetic sequence 21,13,5,3... in the standard form: a_n= a_n=a_1+(n1)dso a_n=21+(n1)8 *a_n=1688n why isnt this right? 
calculus
A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10? 
Calculus
Find a series ∑a_n for which ∑(a_n)^2 converges but ∑a_n diverges 
Calculus
If a_n >0 and b_n >0 and series ∑ sqrt( (a_n)^2 +(b_n)^2 ) converges, then ∑a_n and ∑b_n both converge. True or false? 
Calculus
If a_n>0 and a_(n+1) <= a_n, does the alternating series ∑ ((1)^(n+1)) a_n converge or diverge? 
Algebra
find the arithmetic mean A_n1_3.9, A_n+1_=7.1 
mathematical statistics
Suppose a_n∈ [0,1] and X_n is a sequence of i.i.d random variables with p.d.f : p(X_n=1)= p(X_n= 1)=0.5 . ∑_(n=1)^∞▒a_n X_n is convergent with probability 1, is ∑_(n=1)^∞▒a_n^2 convergent? 
Math Proof
0<=b_n<=a_n. a) if a_n>0 then b_n>0. b) if a_n>infinity then b_n>infinity I don't know where to start. 
Algebra
For the following sequences determine the term indicated: a_1=2, a_n=2(a_n1)^2,a_4 a_n=ln(e^n+2), a_5 b_0=1, b_1=2, b_n+1=2b_nb_0 
DISCRETE MATH
Determine whether the following is a recursive or explicit. Then, find the first four terms of the following sequence. a) a_n = 〖na〗_(n1) where a_0 =5 b) a_n = a_(n1) + 3a_(n2) where a_0 = 1 and a_1 =2 c) a_n = 2^n …